Python Sympy Subset.rank_lexicographic()用法及代碼示例

Subset.rank_lexicographic()：rank_lexicographic()是一個sympy Python庫函數，它返回該子集的字典順序。

用法：

sympy.combinatorics.subset.Subset.rank_lexicographic()

返回：
該子集的字典順序。

代碼1：rank_lexicographic()示例

# Python code explaining 
# SymPy.Subset.rank_lexicographic() 
  
# importing SymPy libraries 
from sympy.combinatorics import Permutation, Cycle 
from sympy.combinatorics.subsets import Subset 
  
# Using from  
# sympy.combinatorics.subset.Subsets.rank_lexicographic() 
  
# Creating Subset 
a = Subset(['c', 'd'], ['geek', 'for', 'c', 'd']) 
  
# rank_lexicographic value 
print ("Subset a rank_lexicographic : ", a.rank_lexicographic)

輸出：

Subset a rank_lexicographic : 14

代碼2：rank_lexicographic()示例

# Python code explaining 
# SymPy.Subset.rank_lexicographic() 
  
# importing SymPy libraries 
from sympy.combinatorics import Permutation, Cycle 
from sympy.combinatorics.subsets import Subset 
  
# Using from  
# sympy.combinatorics.subset.Subsets.rank_lexicographic() 
  
# Creating Subset 
a = Subset([2, 3, 4], [1, 2, 3, 4, 5]) 
  
# rank_lexicographic value 
print ("Subset a rank_lexicographic : ", a.rank_lexicographic)

輸出：

Subset a rank_lexicographic : 19

相關用法

注：本文由純淨天空篩選整理自noobestars101大神的英文原創作品 SymPy | Subset.rank_lexicographic() in Python。非經特殊聲明，原始代碼版權歸原作者所有，本譯文未經允許或授權，請勿轉載或複製。